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I have came across this problem and after trying to answer it for some time, I thought my solution was correct but apparently it is not. Can you please explain to me what I have done wrong ?

Problem:

A train traveling from Aytown to Beetown meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at four-fifths its usual rate, arriving at Beetown 2 hours late. If the train had covered 80 miles before the accident, it would have been just one hour late. What is the usual rate of the train ?

My Attempt:

Let $d$ be the total distance of the trip, $t$ the usual time and $x$ the usual rate. We know that $$x\cdot t = d$$

The first part of the question tells us that $$x\cdot 1+ \frac 45 x(t_2) = d$$ where $t_2$ is the time traveled at four-fifths the usual speed and $$t_2 = (t+2)-\frac 12 - 1 = t+\frac 12.$$ The $(t+2)$ is the "two hours late" part and the $(-1-\frac 12)$ is the break and the one hour before the accident. The second part tells us that $$80 + \frac 45 x(t_3) = d$$ where, again, $t_3$ is the time traveled at four-fifths the usual speed and $$t_3 = (t+1)-\frac 12 - \frac{80}x.$$ Similar to the first equation, $(t+1)$ is the one-hour late part and the $(-\frac 12 - \frac{80}x)$ is the break along with the time spent travelling before the accident. Solving for $x$ I got $16$, however, the answer is $20$. What have I done wrong ?

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    $\begingroup$ Just to clarify, this problem is from "The Art of Problem Solving: Volume 1". $\endgroup$ – Adam Apr 2 '18 at 4:02
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The difference of answer is due to you have answered a different question. You have answered your question correctly. Good job.

This is the question from the book:

A train traveling from Aytown to Beetown meets with an accident after $1$ hour. The train is stopped for $30$ minutes, after which it proceeds at four-fifths its usual rate, arriving at Beetown $2$ hours late. If the train had covered $80$ miles $\color{blue}{\text{more}}$ before the accident, it would have been just one hour late. What is the usual rate of the train ?

In that case, the $80$ miles that are being travelled using different speeds is responsible for the difference of $1$ hour.

$$\frac{80}{4x/5}-\frac{80}{x} =1$$

$$\frac{80}{x}\left(\frac14 \right)=1$$

$$x=20$$

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  • $\begingroup$ Oh wow that's really subtle! $\endgroup$ – Adam Apr 2 '18 at 10:21
  • $\begingroup$ I always miss those little things, it's really frustrating! $\endgroup$ – Adam Apr 2 '18 at 10:22
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I think you got the right answer. You should have continued, solving for the other unknowns, so you could check. I get $x=16, d=112,$ so that the usual trip time is $7$ hours, and $4/5$ of the normal speed is $12.8$ mph.

If the accident occurs after one hour, the train has traveled $16$ miles and has $96$ miles to go, which will take $7.5$ hours at $12.8$ mph. So the length of the trip is $1 + .5+7.5=9,$ two hours longer than usual.

If the accident occurs after $80$ miles, the train has traveled for $5$ hours, and has $32$ miles to go, which will take $2.5$ hours at $12.8$ mph. The total trip time is $5+.5+2.5=8,$ an hour longer than usual.

The $20$ in the book appears to be an error.

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  • $\begingroup$ Oh, thanks a lot! $\endgroup$ – Adam Apr 2 '18 at 4:51
  • $\begingroup$ It was my pleasure. $\endgroup$ – saulspatz Apr 2 '18 at 4:51

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